Excess barium nitrate solution is added to 200 mL of 0.200 mol L⁻¹ sodium sulfate - HSC - SSCE Chemistry - Question 19 - 2016 - Paper 1

The molarity (C) of sodium sulfate is given as 0.200 mol L⁻¹ and the volume (V) is 200 mL, which is equivalent to 0.200 L:

$n_{Na_2SO_4} = C \times V = 0.200 \, \text{mol L}^{-1} \times 0.200 \, \text{L} = 0.0400 \, \text{mol}$

Since the reaction between sodium sulfate and barium nitrate is a 1:1 ratio, the moles of barium sulfate formed will also be 0.0400 mol.

The molar mass of barium sulfate (BaSO₄) is calculated as:

• Molar mass of Ba = 137.33 g/mol
• Molar mass of S = 32.07 g/mol
• Molar mass of O = 16.00 g/mol (4 O atoms)

$\text{Molar mass of BaSO}_4 = 137.33 + 32.07 + (4 \times 16.00) = 233.39 \, \text{g/mol}$

Now, using the number of moles to find the mass:

$\text{mass} = n \times \text{molar mass} = 0.0400 \, \text{mol} \times 233.39 \, \text{g/mol} = 9.34 \, \text{g}$

Based on the calculations, the mass of the solid formed, which is barium sulfate, is approximately 9.34 g. Therefore, the correct answer is (C) 9.33 g.